Paradichlorobenzene, a common ingredient in solid air fresheners, contains only c, h, and cl and has a molar mass of about 147 grams. given that combustion of 1.68 grams of this compound produces 3.02 grams co2 and 0.412 grams h2o, determine its empirical and molecular formulas.

Empirical formula

C3H2Cl

Molecular formula

C6H4Cl2

Explanation:

Firstly, we can get the mass of carbon and hydrogen produced by calculating their number of moles from that of carbon iv oxide and water respectively.

To get the number of moles of carbon iv oxide, we simply make a division. The number of moles is the mass of carbon iv oxide divided by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol. The number of moles is thus 3.02/44 = 0.069

One atom of carbon is present in Carbon iv oxide. This means 0.069 moles of carbon is present. The mass of carbon present equals the atomic mass of carbon multiplied by the number of moles = 0.069 * 12 = 0.83g

The mass of water is 0.412g, the number of moles is 0.412/18 = 0.023

In water, there are two atoms of hydrogen. Hence the number of moles of hydrogen is 2 * 0.023 = 0.046

The mass of hydrogen is the number of moles of hydrogen multiplied by the atomic mass unit of hydrogen = 0.046 * 1 = 0.046g

Now we need to know the mass of chlorine. We simply subtract the masses of hydrogen and carbon.

1.68-0.046-0.83 = 0.804

Now we proceed to calculate the empirical formula. We simply divide the mass of each by their atomic masses and then proceed from there.

H = 0.046/1 = 0.046

C = 0.83/12 = 0.069

Cl = 0.804/35.5 = 0.023

We divide by the smallest number which is that of that of carbon

H = 0.046/0.023= 2

C = 0.069/0.023= 3

Cl = 0.023/0.023 = 1

The empirical formula would thus be C3H2Cl

The molecular formula can be calculated using the molar mass;

(C3H2Cl)n = 147

(36+ 2 + 35.5 ) = 147

73.5n = 147

n = 147/73.5 = 2

The molecular formula is thus C6H4Cl2

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.